Cryptography - Sage

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sage: X = pt.frequency_distribution()sage: m = 11sage: r = 0.75sage: match = [ [] for i in range(m) ]sage: for i in range(m):... Z = ct[i::m].frequency_distribution()... Y = DiscreteRandomVariable(X,Z.function())... for j in range(26):... K = S([ (j+k)%26 for i in range(26) ])... corr = X.translation_correlation(Y,E(K))... if corr > r:... match[i].append(j)Solution. We have already surmised that the first sample ciphertext, cipher01.txt, isoutput from a Vigenére cipher of period 11. We the definitions as below:sage: S = AlphabeticStrings()sage: E = SubstitutionCryptosystem(S)sage: pt = S.encoding(open("Plaintext/blackcat.txt").read())sage: ct = S.encoding(open("Ciphertext/cipher01.txt").read())the output of the above code givessage: match[[], [7], [0], [], [], [18], [3], [4], [0], [], [4]]A translation by 7 corresponds to the character H, by 0 to A, by 18 to S, and by 3 and 4to Dand E, respectively. This gives the partial enciphering key *HA**SDEA*E. Decipheringwith respect to this key gives the plaintext blocks *HE**OETI*I, *KT**NPOM*N, etc.Relaxing the bound from r = 0.75 to 0.50, one finds multiple solutions among them thecorrect solution SHAKESPEARE, giving the plaintextWHENMOSTIWINKTHENDOMINEEYESBESTSEEFORALLTHEDAYTHEYVIEWTHINGSUNRESPWhy is this a bad key choice?Exercise 3.7 (Breaking substitution ciphers) Suppose that rather than an affinetranslation, you have reduced to an arbitrary simple substitution. We need to undo an109
arbitrary permutation of the alphabet.space:For this purpose we define maps into Euclidean1. A → A 2 → R 2 defined byx ↦−→ xx ↦−→ ( P (x), P (xx) ) .2. A → A 2 → R 3 defined byfor some fixed character y.x ↦−→ xy ↦−→ ( P (x), P (xy | y), P (yx | y) ) ,See the documenthttp://echidna.maths.usyd.edu.au/ kohel/tch/Crypto/digraph frequencies.pdffor standard vectors for the English language.Solution. These maps can be applied to the solution of substitution ciphers by findingnearest elements to a known standard for the English language. For instance, assume thatthe ciphertext image x of E has been identified, one can look for pairs xy which are theimage of the plaintext pair ER, by searching for a nearest vector to:(0.05674, 0.13071, 0.11352).This will determine the ciphertext image y of R. This bootstrapping procedure successivelydetermines the substitution from the digraph frequencies of the ciphertext.Exercise 3.8 (Breaking transposition ciphers) In order to break transposition ciphersit is necessary to find the period m, of the cipher, and then to identify positionsi and j within each block 1 + km ≤ i, j ≤ (k + 1)m which were adjacent prior to the permutationof positions. Suppose we guess that m is the correct period. Then for a ciphertextsample C = c 1 c 2 . . . , and a choice of 1 ≤ i < j ≤ m, we can form the digraph decimationsequence c i c j , c i+m c j+m , c i+2m c j+2m , . . . .Two statistical measures that we can use on ciphertext to determine if a digraph sequenceis typical of the English language are a digraph coincidence indexn∑n∑x∈A y∈An xy (n xy − 1)N(N − 1)110 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
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-------------------------------
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: analysis of the each of the decimat
Page 117 and 118: In order to understand naturally oc
Page 119 and 120: We do this by first verifying the e
Page 121 and 122: Solution.None provided.Linear feedb
Page 123 and 124: Multiplying each through by the con
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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