1. Computer Science and Engineering Computer System Security CSE 5339/7339 Lecture 7 September 9, 2004

2. Computer Science and Engineering Contents  Assignment 1   Assignment 2  Implementation of DES  Data Encryption Standard (DES)  Advanced Encryption Standard (AES)  Diana’s Presentation

3. Computer Science and Engineering Group Work Find keys d and e for the RSA cryptosystem with p = 7 and q = 11.

4. Computer Science and Engineering A High Level Description of DES Input - P 16 Cycles Output - C Key IP Inverse IP

5. Computer Science and Engineering A Cycle in DES Right halfLeft half Key shifted And Permuted New R-halfNew L-half f

6. Computer Science and Engineering K 64 bits PC-1 K+ 56 bits C0 28 bitsD0 28 bits C1 28 bits D1 28 bits C2 28 bits D2 28 bits C16 28 bits D16 28 bits PC-2 K1 48 bitsK2 48 bitsK16 48 bits Shift Key Summary

7. Computer Science and Engineering M 64 bits I-P L0 32 bitsR0 32 bits Cycle 1 IP 64 bits f L1 32 bitsR1 32 bits K1 48 bits

8. Computer Science and Engineering L1 32 bitsR1 32 bits Cycle 2 f L2 32 bitsR2 32 bits K2 48 bits

9. Computer Science and Engineering L2 32 bitsR2 32 bits Cycle 3 f L3 32 bitsR3 32 bits K3 48 bits

10. Computer Science and Engineering L15 32 bitsR15 32 bits Cycle 16 f L16 32 bitsR16 32 bits K16 48 bits IP -1 C 64 bits L16 32 bitsR16 32 bits

11. Computer Science and Engineering 32 bits Rn-1 32 bits f Kn 48 bits E E(Rn-1) 48 bits E(Rn-1)+Kn 48 bits S Boxes P

12. Computer Science and Engineering Detailed DES Example (by J. Orlin Grabbe) Plain text message M M = 0123456789ABCDEF (hexadecimal format) M in binary format: M = 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 Left Half (L) and Right Half (R) L = 0000 0001 0010 0011 0100 0101 0110 0111 R = 1000 1001 1010 1011 1100 1101 1110 1111

13. Computer Science and Engineering Key Key K K = K = 133457799BBCDFF1 (hexadecimal format) K in binary format: K = 00010011 00110100 01010111 01111001 10011011 10111100 11011111 11110001 Note: DES operates on the 64-bit blocks using key sizes of 56- bits. The keys are actually stored as being 64 bits long, but every 8th bit in the key is not used (i.e. bits numbered 8, 16, 24, 32, 40, 48, 56, and 64).

14. Computer Science and Engineering Step 1: Create 16 sub-keys (48-bits) 1.1 The 64-bit key is permuted according to table PC-1. 5749413325179 1585042342618 1025951433527 1911360524436 63554739312315 7625446383022 1466153453729 211352820124

15. Computer Science and Engineering Example (cont.) From the original 64-bit key K = 00010011 00110100 01010111 01111001 10011011 10111100 11011111 11110001 Using PC-1, we get the 56-bit permutation K+ = 1111000 0110011 0010101 0101111 0101010 1011001 1001111 0001111

16. Computer Science and Engineering 1.2 Split this key into left and right halves, C 0 and D 0, where each half has 28 bits K+ = 1111000 0110011 0010101 0101111 0101010 1011001 1001111 0001111 From the permuted key K+, we get C 0 = 1111000 0110011 0010101 0101111 D 0 = 0101010 1011001 1001111 0001111

17. Computer Science and Engineering 1.3 Create 16 blocks C n and D n, 1<=n<=16. C n and D n are obtained from C n-1 and D n-1 using the following schedule of "left shifts". 12345678910111213141516 shift 1122222212222221

18. Computer Science and Engineering Example (Cont.) C 0 = 1111000 0110011 0010101 0101111 D 0 = 0101010 1011001 1001111 0001111 C 1 = 1110000110011001010101011111 D 1 = 1010101011001100111100011110 C 2 = 1100001100110010101010111111 D 2 = 0101010110011001111000111101 C 3 = 0000110011001010101011111111 D 3 = 0101011001100111100011110101

19. Computer Science and Engineering Example (Cont.) C 4 = 0011001100101010101111111100 D 4 = 0101100110011110001111010101 C 5 = 1100110010101010111111110000 D 5 = 0110011001111000111101010101 C 6 = 0011001010101011111111000011 D 6 = 1001100111100011110101010101 C 7 = 1100101010101111111100001100 D 7 = 0110011110001111010101010110

20. Computer Science and Engineering Example (Cont.) C 8 = 0010101010111111110000110011 D 8 = 1001111000111101010101011001 C 9 = 0101010101111111100001100110 D 9 = 0011110001111010101010110011 C 10 = 0101010111111110000110011001 D 10 = 1111000111101010101011001100 C 11 = 0101011111111000011001100101 D 11 = 1100011110101010101100110011

21. Computer Science and Engineering Example (Cont.) C 12 = 0101111111100001100110010101 D 12 = 0001111010101010110011001111 C 13 = 0111111110000110011001010101 D 13 = 0111101010101011001100111100 C 14 = 1111111000011001100101010101 D 14 = 1110101010101100110011110001 C 15 = 1111100001100110010101010111 D 15 = 1010101010110011001111000111

22. Computer Science and Engineering 1.4 Form the keys K n, for 1<=n<=16, by applying the following permutation table to each of the concatenated pairs C n D n. Each pair has 56 bits, but PC-2 only uses 48 of these. 1417112415 3281562110 2319124268 1672720132 415231374755 304051453348 444939563453 464250362932

23. Computer Science and Engineering Example (Cont.) For the first key we have C 1 D 1 = 1110000 1100110 0101010 1011111 1010101 0110011 0011110 0011110 which, after we apply the permutation PC-2, becomes K 1 = 000110 110000 001011 101111 111111 000111 000001 110010

24. Computer Science and Engineering Example (Cont.) K 2 = 011110 011010 111011 011001 110110 111100 100111 100101 K 3 = 010101 011111 110010 001010 010000 101100 111110 011001 K 4 = 011100 101010 110111 010110 110110 110011 010100 011101 K 5 = 011111 001110 110000 000111 111010 110101 001110 101000 K 6 = 011000 111010 010100 111110 010100 000111 101100 101111 K 7 = 111011 001000 010010 110111 111101 100001 100010 111100

25. Computer Science and Engineering Example (Cont.) K 8 = 111101 111000 101000 111010 110000 010011 101111 111011 K 9 = 111000 001101 101111 101011 111011 011110 011110 000001 K 10 = 101100 011111 001101 000111 101110 100100 011001 001111 K 11 = 001000 010101 111111 010011 110111 101101 001110 000110 K 12 = 011101 010111 000111 110101 100101 000110 011111 101001

26. Computer Science and Engineering Example (Cont.) K 13 = 100101 111100 010111 010001 111110 101011 101001 000001 K 14 = 010111 110100 001110 110111 111100 101110 011100 111010 K 15 = 101111 111001 000110 001101 001111 010011 111100 001010 K 16 = 110010 110011 110110 001011 000011 100001 011111 110101

27. Computer Science and Engineering Step 2: Encode each 64-bit block of data 2.1 Do initial permutation IP of M to the following IP table. 585042342618102 605244362820124 625446383022146 645648403224168 57494133251791 595143352719113 615345372921135 635547393123157

28. Computer Science and Engineering Example (Cont.) Applying the initial permutation to the block of text M, we get M = 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 IP = 1100 1100 0000 0000 1100 1100 1111 1111 1111 0000 1010 1010 1111 0000 1010 1010

29. Computer Science and Engineering 2.2 Divide the permuted block IP into a left half L 0 of 32 bits, and a right half R 0 of 32 bits IP = 1100 1100 0000 0000 1100 1100 1111 1111 1111 0000 1010 1010 1111 0000 1010 1010 From IP we get L 0 = 1100 1100 0000 0000 1100 1100 1111 1111 R 0 = 1111 0000 1010 1010 1111 0000 1010 1010

30. Computer Science and Engineering 2.3 Proceed through 16 iterations, for 1<=n<=16, using a function f which operates on two blocks— a data block of 32 bits and a key K n of 48 bits to produce a block of 32 bits. L n = R n-1 R n = L n-1 + f(R n-1,K n ) -- + denote XOR K 1 = 000110 110000 001011 101111 111111 000111 000001 110010 L 1 = R 0 = 1111 0000 1010 1010 1111 0000 1010 1010 R 1 = L 0 + f(R 0,K 1 )

31. Computer Science and Engineering The Calculation of the function f 1 - Expand R n-1  E(R n-1 ) 2- XOR  K n + E(R n-1 ) = B 1 B 2 B 3 B 4 B 5 B 6 B 7 B 8 3- Substitution S-Boxes  S 1 (B 1 )S 2 (B 2 )S 3 (B 3 )S 4 (B 4 )S 5 (B 5 )S 6 (B 6 )S 7 (B 7 )S 8 (B 8 ) 4- P permutation  f = P(S 1 (B 1 )S 2 (B 2 )...S 8 (B 8 ))

32. Computer Science and Engineering 2.4 Expand each block R n-1 from 32 bits to 48 bits using a selection table that repeats some of the bits in R n-1. 3212345 456789 8910111213 121314151617 161718192021 202122232425 242526272829 28293031321

33. Computer Science and Engineering We'll call the use of this selection table the function E. Thus E(R n-1 ) has a 32 bit input block, and a 48 bit output block. E R n-1 E(R n-1 )

34. Computer Science and Engineering Example (Cont.) We calculate E(R 0 ) from R 0 as follows: R 0 = 1111 0000 1010 1010 1111 0000 1010 1010 E(R 0 ) = 011110 100001 010101 010101 011110 100001 010101 010101 Note that each block of 4 original bits has been expanded to a block of 6 output bits.

35. Computer Science and Engineering XOR Operation In the f calculation, we XOR the output E(R n-1 ) with the key K n : K n + E(R n-1 ) K 1 = 000110 110000 001011 101111 111111 000111 000001 110010 E(R 0 ) = 011110 100001 010101 010101 011110 100001 010101 010101 K 1 +E(R 0 ) = 011000 010001 011110 111010 100001 100110 010100 100111.

36. Computer Science and Engineering Substitution – S-Boxes We now have 48 bits, or eight groups of six bits. We use each group of 6 bits as addresses in tables called "S boxes". Each group of six bits will give us an address in a different S box. Located at that address will be a 4 bit number. This 4 bit number will replace the original 6 bits. The net result is that the eight groups of 6 bits are transformed into eight groups of 4 bits (the 4-bit outputs from the S boxes) for 32 bits total.

37. Computer Science and Engineering Substitution – S-Boxes (Cont.) K n + E(R n-1 ) = B 1 B 2 B 3 B 4 B 5 B 6 B 7 B 8 where each B i is a group of six bits. We now calculate S 1 (B 1 )S 2 (B 2 )S 3 (B 3 )S 4 (B 4 )S 5 (B 5 )S 6 (B 6 )S 7 (B 7 )S 8 (B 8 ) where S i (B i ) referrers to the output of the i-th S box.

38. Computer Science and Engineering Substitution – S-Boxes (Cont.) 0123456789101112131415 01441312151183106125907 10157414213110612119539 24114813621115129731050 31512824917511314100613 Box S1

39. Computer Science and Engineering Finding S 1 (B 1 ) The first and last bits of B represent in base 2 a number in the decimal range 0 to 3. Let that number be i. The middle 4 bits of B represent in base 2 a number in the decimal range 0 to 15. Let that number be j. Look up in the table the number in the i-th row and j-th column. The tables defining the functions S 1,...,S 8 are given in page 652

40. Computer Science and Engineering Example (Cont.) For input block B = 011011 the first bit is "0" and the last bit "1" giving 01 as the row. This is row 1. The middle four bits are "1101". This is the binary equivalent of decimal 13, so the column is column number 13. In row 1, column 13 appears 5. This determines the output; 5 is binary 0101, so that the output is 0101. Hence S 1 (011011) = 0101.

41. Computer Science and Engineering Example (Cont.) For the first round, we obtain as the output of the eight S boxes: K 1 + E(R 0 ) = 011000 010001 011110 111010 100001 100110 010100 100111. S 1 (B 1 )S 2 (B 2 )S 3 (B 3 )S 4 (B 4 )S 5 (B 5 )S 6 (B 6 )S 7 (B 7 )S 8 (B 8 ) = 0101 1100 1000 0010 1011 0101 1001 0111 For the first round, we obtain as the output of the eight S boxes: K 1 + E(R 0 ) = 011000 010001 011110 111010 100001 100110 010100 100111. S 1 (B 1 )S 2 (B 2 )S 3 (B 3 )S 4 (B 4 )S 5 (B 5 )S 6 (B 6 )S 7 (B 7 )S 8 (B 8 ) = 0101 1100 1000 0010 1011 0101 1001 0111

42. Computer Science and Engineering Permutation P of the S-box output f = P(S 1 (B 1 )S 2 (B 2 )...S 8 (B 8 )) 1672021 29122817 1152326 5183110 282414 322739 1913306 2211425

43. Computer Science and Engineering Example (Cont.) For the first round, we obtain as the output of the eight S boxes: K 1 + E(R 0 ) = 011000 010001 011110 111010 100001 100110 010100 100111. S 1 (B 1 )S 2 (B 2 )S 3 (B 3 )S 4 (B 4 )S 5 (B 5 )S 6 (B 6 )S 7 (B 7 )S 8 (B 8 ) = 0101 1100 1000 0010 1011 0101 1001 0111 From the output of the eight S boxes: S 1 (B 1 )S 2 (B 2 )S 3 (B 3 )S 4 (B 4 )S 5 (B 5 )S 6 (B 6 )S 7 (B 7 )S 8 (B 8 ) = 0101 1100 1000 0010 1011 0101 1001 0111 we get f = 0010 0011 0100 1010 1010 1001 1011 1011

44. Computer Science and Engineering Example (Cont.) For the first round, we obtain as the output of the eight S boxes: K 1 + E(R 0 ) = 011000 010001 011110 111010 100001 100110 010100 100111. S 1 (B 1 )S 2 (B 2 )S 3 (B 3 )S 4 (B 4 )S 5 (B 5 )S 6 (B 6 )S 7 (B 7 )S 8 (B 8 ) = 0101 1100 1000 0010 1011 0101 1001 0111 R 1 = L 0 + f(R 0, K 1 ) = 1100 1100 0000 0000 1100 1100 1111 1111 + 0010 0011 0100 1010 1010 1001 1011 1011 = 1110 1111 0100 1010 0110 0101 0100 0100

45. Computer Science and Engineering Process Repeated 16 rounds For the first round, we obtain as the output of the eight S boxes: K 1 + E(R 0 ) = 011000 010001 011110 111010 100001 100110 010100 100111. S 1 (B 1 )S 2 (B 2 )S 3 (B 3 )S 4 (B 4 )S 5 (B 5 )S 6 (B 6 )S 7 (B 7 )S 8 (B 8 ) = 0101 1100 1000 0010 1011 0101 1001 0111 In the next round, we will have L 2 = R 1, which is the block we just calculated, and then we must calculate R 2 =L 1 + f(R 1, K 2 ), and so on for 16 rounds.

46. Computer Science and Engineering Final Phase At the end of the sixteenth round we have L 16 and R 16. We then reverse the order of the two blocks into R 16 L 16 and apply a final permutation IP -1 as defined by the following table 408481656246431 397471555236331 386461454226230 375451353216129 364441252206028 353431151195927 342421050185826 33141949175725

47. Computer Science and Engineering Example (cont.) If we process all 16 blocks using the method defined previously, we get, on the 16th round, L 16 = 0100 0011 0100 0010 0011 0010 0011 0100 R 16 = 0000 1010 0100 1100 1101 1001 1001 0101

48. Computer Science and Engineering Example (cont.) We reverse the order of these two blocks and apply the final permutation to R 16 L 16 = 00001010 01001100 11011001 10010101 01000011 01000010 00110010 00110100 IP -1 = 10000101 11101000 00010011 01010100 00001111 00001010 10110100 00000101, which in hexadecimal format is 85E813540F0AB405

49. Computer Science and Engineering The End M = 0123456789ABCDEF C = 85E813540F0AB405 Decryption is simply the inverse of encryption, following the same steps as above, but reversing the order in which the sub- keys are applied

50. Computer Science and Engineering Triple DES Triple-DES is just DES with two 56-bit keys applied. Given a plaintext message, the first key is used to DES- encrypt the message. The second key is used to DES-decrypt the encrypted message. (Since the second key is not the right key, this decryption just scrambles the data further.) The twice-scrambled message is then encrypted again with the first key to yield the final ciphertext. This three-step procedure is called triple-DES.

51. Computer Science and Engineering Advanced Encryption Standard (ASE) -NIST, call 1997 -One was selected out of five -Rijndael (Rine dahl)  Vincent Rijmen & Joam Daemen -In 2001, it was formally adopted by US -9, 11, 13 cycles (rounds) for keys of 128, 192, 256 bits

52. Computer Science and Engineering ASE (cont) -Each cycle consists of 4 steps - Byte substitution (substitution) - Shift row (transposition) - Mix column - Add subkey

53. Computer Science and Engineering State -128-bit block  4 x 4 matrix -128 bits  b0, b1, b2,.., b15 b0b4b8b12 b1b5b9b13 b2b6b10b14 b3b7b11b15

54. Computer Science and Engineering 4 Operations 1. s[i,j]  s’[i,j] (predefined substitution table, Table 10-11 page 663) 2. Rows – left circular shift 3. The 4 elements in each column are multiplied by a polynomial 4. Key is derived and added to each column

55. Computer Science and Engineering Shift Row 15913 610142 111537 164812 15913 261014 371115 481216

56. Computer Science and Engineering Mix Column 2311 1231 1123 3112 = * Multiplying by 1  no change Multiplying by 2  shift left one bit Multiplying by 3  shift left one bit and XOR with original value More than 8 bits  100011011 is subtracted

57. Computer Science and Engineering Add Subkey Group Exercise