Cryptography - Sage

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LFSR Cryptosystems We introduce new utilities for binary stream cryptosystems basedon linear feedback shift registers. The functions binary encoding and binary decodingconvert ASCII text into its bit sequence and back. In addition, the new binary cryptosystemsare:LFSRCryptosystemShrinkingGeneratorCryptosystemUnlike the encoding function strip encoding we have used so far, the functionbinary encoding is information-preserving, taking 8-bit ASCII input and returning thebinary encoding string. The inverse function binary decoding recovers the original text.A linear feedback shift register cryptosystem is created in SAGE using the functionLFSRCryptosystem, taking no arguments. A key is defined by means of a pair, consistingof the connection polynomial g(x) over F 2 and a initial bit sequence of length equal to thedegree of the sequence. A sample use of the cryptosystem follows. The shrinking generatorcryptosystem is a cryptosystem based on a pair of LFSR’s as defined in class.sage: F2 = FiniteField(2)sage: P2. = PolynomialRing(F2)sage: g = x^17 + x^5 + 1sage: IS = [ F2.random_element() for i in range(17) ]sage: LFSR = LFSRCryptosystem()sage: PT = LFSR.encoding("The dog ate my assignment."); PT0101010001101000011001010010000001100100011011110110011100100000011000010111010001100101001000000110110101111001001000000110000101110011011100110110100101100111011011100110110101100101011011100111010000101110sage: K = (g,IS)sage: e = LFSR(K)sage: CT = e(PT)sage: PT == e(CT)TrueNote that the encoding of the message is not ciphertext – this is the standard ASCII bitencoding.Exercise 6.7 Consider the coefficient sequence for f(x)/g(x) in F 2 [[x]], where g(x) =1 + x + x 4 and f(x) = 1 + x 3 . Is g(x) an irreducible polynomial? A primitive polynomial?Draw the associated linear feedback shift register. What is the initial state of the shiftregister?Exercise 6.8 Compute the linear complexity of the sequences 11, 1011, 10101, 10110, and10011.Linear Complexity 57
Exercise 6.9 Compute the first 8 terms of the linear complexity profile of the coefficientsequence from Exercise 1.Exercise 6.10 Practice encoding and enciphering with the LFSR stream cryptosystem.The function binary decoding easily converts this back to ASCII text. Use these functionsto verify that PT is just the binary encoding of the original plaintext message and that theciphertext is enciphered.Exercise 6.11 Since the LFSR is the bitsum of the binary keystream, generated by theconnection polynomial and initial state, why must the inverse key be equal to the key itself?58 Chapter 6. Stream Ciphers
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Author (David R. Kohel) /Title (Cry
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CONTENTS1 Introduction to Cryptogra
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PrefaceWhen embarking on a project
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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 44 and 45: Deciphering. Suppose we begin with
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: Exercise 6.6 In the previous exerci
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-------------------------------
Page 95 and 96: sage: x.is_unit?Type:builtin_functi
Page 97 and 98: Python (hence SAGE) has useful data
Page 99 and 100: sage: n = 12sage: for i in range(n)
Page 101 and 102: sage: I = [55+i for i in range(3)]
Page 103 and 104: sage: I = [7, 4, 11, 11, 14, 22, 14
Page 105 and 106: ExercisesRead over the above SAGE t
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Page 109 and 110: Solution. The block length is the n
Page 111 and 112: 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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