Cryptography - Sage

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APPENDIXDRevision ExercisesLet A be the alphabet {A, B, C, D, E}. Given the message A BAD CAB A DEAD DAD, we formthe strip–encoded plaintextM = ABADCABADEADDADby removing all characters not in the alphabet.Exercise D.1 Encipher the message M using the substitution key K = BDEAC. Find theinverse key and verify the correctness by deciphering your ciphertext.Solution. Recall that the key K = BDEAC specifies the map A ↦→ B, B ↦→ D, etc. Thisresults in the encipheringM = ABADCABADEADDAD ↦→ C = BDBAEBDBACBAABD.Exercise D.2 Let A → Z/5Z be the bijection A ↦→ 0, B ↦→ 1, . . . , E ↦→ 4. Encipher themessage M using the Vigenère key K = ADECB in ECB mode, then encipher the samemessage using the same key and initialization vector BBBB, in CFB and OFB modes withthe block length n = 5 and r = 1. Rather than bit sum, use summation in Z/5Z for thefeedback. Verify the correctness of your results by then deciphering the ciphertext.Solution. We make the identification M = ABADCABADEADDAD = 010320103403303 overZ/5Z. In the same way, we write K = ADECB = 03421. The enciphering in ECB mode isthen 044030440001223 = AEEADAEEAAABCCD. The enciphering in CBC mode begins withC 0 = BBBBB = 11111, and since E K (C j−1 ⊕ M j ) = C j−1 ⊕ E K (M j ), we just add in theprevious ciphertext block to get11111100141441410132 = BBBBBBAABEBEEBEBABDC.133
The application of this function E K to form the state vectors I j is particularly weak, sinceonly the first character of the key K and the first character of the initialization vector.Since K = A ∗ ∗ ∗ ∗, this means I j = B ∗ ∗ ∗ ∗, and so the ciphertext output isBBBBBBCBEDBCBEABEEBE.Exercise D.3 Let K = [3, 5, 4, 1, 2] be a transposition key. Encipher the message M inECB mode and in CBC mode. Verify the correctness of your results by deciphering theciphertext.Solution. The key K = [3, 5, 4, 1, 2] specifies a transposition, under which M ↦→ACDABAEDABDDAAD in ECB mode. If we use the addition ⊕ of the previous question for thefeedback, then we get ciphertextBBBBBACDABDADADCCAAC.Check this work carefully for errors – no guarantees.Exercise D.4 Which of the modes of operation leaves Vigenère ciphertext open to attackby the Kasiski method? Which mode of operation was used for the block ciphers in thecourse assignments, and why?Solution. We made use of the ECB mode in order to preserve the structure of a Vigenèreciphertext. This leaves this and other classical cryptosystems open to classical attacks suchas the Kasiski method.Mathematics of LFSR’sNext we focus on some of the mathmematical problems which arise in stream ciphers andpublic key cryptography. The problems given are of a size which can be computed by hand,with minimal effort if the proper method is used.Exercise D.5 Let S be the set {x 6 +x+1, x 6 +x 3 +1, x 6 +x 5 +1, x 6 +x 2 +1} of polynomialsin F 2 [x].1. Which of the polynomials are irreducible?2. Which of the polynomials are primitive?3. What are the periods of the linear feedback shift registers with the above connectionspolynomials?134 Appendix D. Revision 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
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CHAPTEREIGHTPublic Key Cryptography
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Initial setup:1. Alice and Bob publ
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We apply this rule in the RSA algor
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the discrete logarithm problem (DLP
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Man in the Middle AttackThe man-in-
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Exercise 8.6 Fermat’s little theo
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k < p − 1 with GCD(k, p − 1) =
Page 88 and 89: CHAPTERTENSecret SharingA secret sh
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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 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: sage: a5 := a^n5sage: c5 := c^n5sag
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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