Cryptography - Sage

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Deciphering. Suppose we begin with C = R r R r−1 = R ′ −1R ′ 0, and a reversed key sequenceK ′ 1, K ′ 2 . . . , K ′ r = K r , K r−1 . . . , K 1 . The deciphering follows the same algorithm asenciphering with respect to this key sequence:R ′ j+1 = R ′ j−1 ⊕ f K ′j+1(R ′ j). (5.2)Setting j = r − i − 1, we have K ′ j+1 = K ′ r−i = K i+1 . We moreover want to show therelationsR ′ −1 = R r , R ′ 0 = R r−1 , . . . , R ′ r−1 = R 0 , R ′ r = R −1 .In other words, we want to show that R ′ j = R i whenever i + j = r − 1.Clearly this relation holds for (i, j) = (r, −1) and (i, j) = (r − 1, 0). Assuming it holdsfor j − 1 and j we prove that it holds for j + 1. The deciphering sequence (5.2) can bereplaced byR ′ j+1 = R ′ j−1 ⊕ f K ′j+1(R ′ j) = R ′ r−i−2 ⊕ f K ′r−i(R ′ r−i−1) = R i+1 ⊕ f Ki+1 (R i )The expression R i+1 = R i−1 ⊕f Ki+1 (R i ) in (5.2) can be rearranged by adding (= substracting)f Ki+1 (R i ) to both sides to get R i+1 ⊕f Ki+1 (R i ) = R i−1 . We conclude that R ′ j+1 = R i−1 ,so the equality holds by induction.Example 5.1 (Feistel cipher) Let f Kicomposition of the following maps:be the block cipher, of block length 4, which is the1. The transposition cipher T = [4, 2, 1, 3]; followed by2. A bit-sum with the 4-bit key K i ; followed by3. A substitution cipher S applied to the 2-bit blocksS(00) = 10, S(10) = 01, S(01) = 11, S(11) = 00,i.e. b 1 b 2 b 3 b 4 ↦→ S(b 1 b 2 )S(b 3 b 4 ).Let C be the 3-round Feistel cryptosystem of key length 12, where the three internal keysK 1 , K 2 , K 3 are the first, second, and third parts of the input key K, and the round functionis f Ki .Exercise. Compute the enciphering of the text M = 11010100, using the key K =001011110011.Feistel ciphers 39
5.2 Digital Encryption Standard OverviewThe DES is a 16-round Feistel cipher, which is preceeded and followed by an initial permutationIP and its inverse IP −1 . That is, we start with a message M, and take L 0 R 0 = IP (M)as input to the Feistel cipher, with output IP −1 (R 16 L 16 ). The 64-bits of the key are usedto generate 16 internal keys, each of 48 bits. The steps of the round function f K is givenby the following sequence, taking on 32-bit strings, expanding them to 48-bit strings, andapplying a 48-bit block function.1. Apply a fixed expansion permutation E — this function is a permutation the 32 bitswith repetitions to generate a 48-bit block E(R i ).2. Compute the bit-sum of E(R i ) with the 48-bit key K i , and write this as 8 blocksB 1 , . . . , B 8 of 6 bits each.3. Apply to each block B j = b 1 b 2 b 3 b 4 b 5 b 6 a substitution S j . These substitutions arespecified by S-boxes, which describe the substitution as a look-up table. The outputof the substitution cipher is a 4-bit string C j , which results in the 32-bit stringC 1 C 2 C 3 C 4 C 5 C 6 C 7 C 8 .4. Apply a fixed 32-bit permutation P to C 1 C 2 C 3 C 4 C 5 C 6 C 7 C 8 , and output the resultas f Ki (R).This completes the description of the round function f Ki .5.3 Advanced Encryption Standard OverviewIn 1997, the NIST called for submissions for a new standard to replace the aging DES.The contest terminated in November 2000 with the selection of the Rijndael cryptosystemas the Advanced Encryption Standard (AES).The Rijndael cryptosystem operates on 128-bit blocks, arranged as 4 × 4 matrices with8-bit entries. The algorithm consists of multiple iterations of a round cipher, each of whichis the composition of the following four basic steps:• ByteSub transformation. This step is a nonlinear substition, given by a S-box (lookuptable), designed to resist linear and differential cryptanalysis.• ShiftRow transformation. Provides a linear mixing for diffusion of plaintext bits.• MixColumn transformation. Provides a similar mixing as in the ShiftRow step.• AddRoundKey transformation. Bitwise XOR with the round key.40 Chapter 5. Block Ciphers
Page 1 and 2: Author (David R. Kohel) /Title (Cry
Page 4 and 5: CONTENTS1 Introduction to Cryptogra
Page 6: PrefaceWhen embarking on a project
Page 10 and 11: information. We introduce here some
Page 12 and 13: ut strings in A ∗ map injectively
Page 14 and 15: CHAPTERTWOClassical Cryptography2.1
Page 16 and 17: LV MJ CW XP QO IG EZ NB YH UA DS RK
Page 18 and 19: As a special case, consider 2-chara
Page 20 and 21: Note that if d k = 1, then we omit
Page 22: ExercisesSubstitution ciphersExerci
Page 25 and 26: Ciphertext-only AttackThe cryptanal
Page 27 and 28: of size n, suppose that p i is the
Page 29 and 30: Note that ZKZ and KZA are substring
Page 31: Checking possible keys, the partial
Page 34 and 35: sage: X = pt.frequency_distribution
Page 36 and 37: CHAPTERFOURInformation TheoryInform
Page 38 and 39: For each of these we can extend our
Page 40 and 41: in terms of the cryptosystem), then
Page 42 and 43: CHAPTERFIVEBlock CiphersData Encryp
Page 46 and 47: The Advanced Encryption Standard al
Page 48 and 49: 1. Malicious substitution of a ciph
Page 50 and 51: locks M j−1 , . . . , M 1 as well
Page 52: where X = K ⊕ M = (X 1 , X 2 , X
Page 55 and 56: 6.2 Properties of Stream CiphersSyn
Page 57 and 58: Exercise. Verify that the equality
Page 59 and 60: n 2 n − 11 12 33 74 155 316 637 1
Page 61 and 62: Exercise 6.6 In the previous exerci
Page 63 and 64: Exercise 6.9 Compute the first 8 te
Page 65 and 66: which holds since −4 = 17 + (−1
Page 67 and 68: must therefore have a divisor of de
Page 69 and 70: Shrinking Generator cryptosystemLet
Page 72 and 73: CHAPTEREIGHTPublic Key Cryptography
Page 74 and 75: Initial setup:1. Alice and Bob publ
Page 76 and 77: We apply this rule in the RSA algor
Page 78 and 79: the discrete logarithm problem (DLP
Page 80 and 81: Man in the Middle AttackThe man-in-
Page 82: Exercise 8.6 Fermat’s little theo
Page 85 and 86: k < p − 1 with GCD(k, p − 1) =
Page 88 and 89: CHAPTERTENSecret SharingA secret sh
Page 90: using any t shares (x 1 , y 1 ), .
Page 93 and 94: sage-------------------------------
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sage: x.is_unit?Type:builtin_functi
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Python (hence SAGE) has useful data
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sage: n = 12sage: for i in range(n)
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sage: I = [55+i for i in range(3)]
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sage: I = [7, 4, 11, 11, 14, 22, 14
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ExercisesRead over the above SAGE t
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102
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Solution. The block length is the n
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Solution.below.The coincidence inde
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analysis of the each of the decimat
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arbitrary permutation of the alphab
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In order to understand naturally oc
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We do this by first verifying the e
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Solution.None provided.Linear feedb
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Multiplying each through by the con
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Solution. The linear complexity of
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If a, b, and c are as above, then f
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Exercise 8.5 Use SAGE to find a lar
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Solution. Now we can verify that e
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which has no common factors with p
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sage: p = 2^32+61sage: m = (p-1).qu
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sage: a5 := a^n5sage: c5 := c^n5sag
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The application of this function E
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5. (∗) How many elements a of G h
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1. The value f(0) of the polynomial
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Cryptography - Sage

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