Cryptography - Sage

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Solution. The data of a LFSR diagram, of a linear recurrences relation, and of a connectionpolynomial are equivalent — they express the same information. The connectionpolynomial g(x) = ∑ j c jx j encodes the wiring of a LFSR which implements a recurrencerelation. Thinking of x j as a shift operator acting on the sequence s 0 , s 1 , s 2 , . . . , the behaviourof g(x) in the product g(x)s(x) (below) is precisely this recurrence relation.Exercise 6.5 For a linear feedback register of length n, define a power series∞∑s(x) = s i x ifrom the output sequence s i . Suppose that the linear feedback register defines the recursions i+n = ∑ n−1j=0 c n−js i+j . Define a polynomial g(x) = 1 + ∑ n−1j=1 c jx j . Show that f(x) =g(x)s(x) is a polynomial, that is, all of its coefficients are eventually zero. What is thepolynomial f(x)?Solution. The expression f(x) = s(x)g(x) for a polynomial f(x) of degree less thann = deg(g(x)) is another equivalent formulation of the recurrence relation. The initialn coefficients of s(x) are the entries of the shift register, and the n coefficients of f(x) isa linear combination of these coefficients. Although the coefficients of f(x) are not equalto the initial state, for a nonsingular LFSR, the initial states are in bijection with thenumerator polynomials f(x).Exercise 6.6 In the previous exercise we showed that the power series s(x) has the formf(x)/g(x) in the power series ring F 2 [[x]]. In SAGE it is possible to form power seriesrings in the following wayi=1sage: F2 = FiniteField(2)sage: PS. = PowerSeriesRing(F2)sage: f = x^2 + xsage: g = x^3 + x + 1sage: f/g + O(x^16)x + x^4 + x^5 + x^6 + x^8 + x^11 + x^12 + x^13 + x^15 + O(x^16)Consider the linear feedback shift register at the beginning of the worksheet. Constructthe corresponding power series and verify that these are the same of the output sequencesthat you computed.Solution.The power series expansions of the first question are:s(x) = x + x 2 + x 5 + x 9 + x 10 + x 11 + x 12 + x 14 + · · ·s(x) = 1 + x + x 2 + x 4 + x 6 + x 7 + x 10 + x 14 + x 15 + · · ·s(x) = 1 + x 2 + x 4 + x 5 + x 8 + x 12 + x 13 + x 14 + x 15 + · · ·s(x) = 1 + x + x 4 + x 8 + x 9 + x 10 + x 11 + x 13 + x 15 + · · ·117
Multiplying each through by the connection polynomial g(x) = x 4 + x + 1, we find thenumerator polynomials for each of the sequences:f(x) = x + x 3f(x) = 1 + x 3f(x) = 1 + x + x 2 + x 3f(x) = 1 + x 2This verifies that the sequences output are consistent with their expected structure ascoefficients of a rational power series.Statistical Properties. The output of a linear feedback shift operator of length n has aperiod, which must divide 2 n − 1. The period is independent of the initial state, providedit is non-zero. If the period equals 2 n − 1, the output sequence is said to be an m-sequence.The following theorem describes the statistical properties of m-sequences.Theorem 6.1 Let s = s 0 s 1 . . . be an m-sequence, and let k be an integer with 1 ≤ k ≤ n.Then in each subsequence of s of length 2 n + k − 2, every finite nonzero binary sequenceof length k appears as a subsequence exactly 2 n−k times, and the length k zero subsequenceappears exactly 2 n−k − 1 times.A polynomial g(x) in F 2 [x] is irreducible if it is not the product of two polynomials ofdegree greater than zero. An irreducible polynomial g(x) of degree n is primitive if g(x)divides x N − 1 for N = 2 n − 1 and no smaller value of N. (Equivalently, g(x) is primitiveif the powers x i are distinct modulo g(x), for i with 1 ≤ i ≤ 2 n − 1.)Linear Complexity. A linear feedback shift register is said to generate a binary sequences if there exists some initial state for which its output sequence is s. The linear complexityL(s) of an infinite sequence s is defined to be zero if s is the zero sequence, infinity if s isgenerated by no linear feedback shift register, and otherwise equal to the minimal length ofa linear feedback shift register generating s. The linear complexity L(s) of a finite sequences is defined to be the minimal length of a linear feedback shift register with initial sequences for some initial state.Linear Complexity Profile. For a sequence s, define L j (s) to be the linear complexityof the first j terms of the sequence. The linear complexity profile of an infinite sequences is defined to be the infinite sequence L 1 (s), L 2 (s), . . . , and for a finite sequence s =s 0 s 1 . . . s N−1 is defined to be the finite sequence L 1 (s), L 2 (s), . . . , L n (s).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:LFSRCryptosystemShrinkingGeneratorCryptosystem118 Appendix C. Solutions to Exercises
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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
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ut strings in A ∗ map injectively
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CHAPTERTWOClassical Cryptography2.1
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LV MJ CW XP QO IG EZ NB YH UA DS RK
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As a special case, consider 2-chara
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Note that if d k = 1, then we omit
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ExercisesSubstitution ciphersExerci
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Ciphertext-only AttackThe cryptanal
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of size n, suppose that p i is the
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Note that ZKZ and KZA are substring
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Checking possible keys, the partial
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sage: X = pt.frequency_distribution
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CHAPTERFOURInformation TheoryInform
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For each of these we can extend our
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in terms of the cryptosystem), then
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CHAPTERFIVEBlock CiphersData Encryp
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Deciphering. Suppose we begin with
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The Advanced Encryption Standard al
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1. Malicious substitution of a ciph
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locks M j−1 , . . . , M 1 as well
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where X = K ⊕ M = (X 1 , X 2 , X
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6.2 Properties of Stream CiphersSyn
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Exercise. Verify that the equality
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n 2 n − 11 12 33 74 155 316 637 1
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Exercise 6.6 In the previous exerci
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Exercise 6.9 Compute the first 8 te
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which holds since −4 = 17 + (−1
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must therefore have a divisor of de
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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
Page 113 and 114: analysis of the each of the decimat
Page 115 and 116: arbitrary permutation of the alphab
Page 117 and 118: In order to understand naturally oc
Page 119 and 120: We do this by first verifying the e
Page 121: Solution.None provided.Linear feedb
Page 125 and 126: Solution. The linear complexity of
Page 127 and 128: If a, b, and c are as above, then f
Page 129 and 130: Exercise 8.5 Use SAGE to find a lar
Page 131 and 132: Solution. Now we can verify that e
Page 133 and 134: which has no common factors with p
Page 135 and 136: sage: p = 2^32+61sage: m = (p-1).qu
Page 137 and 138: sage: a5 := a^n5sage: c5 := c^n5sag
Page 139 and 140: The application of this function E
Page 141 and 142: 5. (∗) How many elements a of G h
Page 143: 1. The value f(0) of the polynomial
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Cryptography - Sage

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